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wrapping parameterized types
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Date: 2007-05-04 (11:47)
From: rossberg@p...
Subject: Re: [Caml-list] wrapping parameterized types
"skaller" <skaller@users.sourceforge.net> wrote:
> On Thu, 2007-05-03 at 19:16 -0400, Chris King wrote:
>> The solution is to use existential types.  In a record, you can tell
>> O'Caml that a particular function _must_ be polymorphic:
>> type 'b mylistfun = { listfun: 'a. 'a list -> 'b }
> I'm still confused why this is called an existential, when
> clearly the quantification is universal.

You have reason to be confused, because this is no existential type.

Dirk Thierbach:
>It's because the universal quantifier is in a "negative" position,
>which is equivalent to an existential quantifier on the outside.
>Just pretend they are logic formulae instead of types, and then
>(\forall a. a) -> b   is equivalent to   \exists a. (a -> b)

Actually, no, these are not equivalent. Only the following are:

(\exists a. a) -> b   is equivalent to   \forall a. (a -> b)

Here is the constructive proof. Assume:

  f : (exists a.a) -> b
  g : forall a. (a -> b)

You can construct g from f and vice versa:

  g = \a. \a:x. f <a,x>
  f = \y:(exists a.a). let <a,x> = y in g a x

- Andreas